Implicit DifferentiationDate: 11/1/96 at 19:12:28 From: brenchi Subject: Implicit Differentiation Given that (x^p)*(y^q) = (x+y)^(p+q), prove that: dy/dx = y/x. Thanks a lot! From Billy Date: 11/2/96 at 16:39:1 From: Doctor Anthony Subject: Re: Implicit Differentiation Starting with (x^p)*(y^q) = (x+y)^(p+q), take logs of both sides: p.ln(x) + q.ln(y) = (p+q)ln(x+y) differentiate implicitly: p/x + (q/y).dy/dx = (p+q)/(x+y)[(1 + dy/dx)] collect terms in dy/dx: dy/dx[q/y - (p+q)/(x+y)] = (p+q)/(x+y) - p/x dy/dx[(qx+qy-py-qy)/y(x+y)] = (px+qx-px-py)/x(x+y) dy/dx (qx-py)/y = (qx-py)/x dy/dx = y/x -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/